Đáp án:
Bạn tham khảo lời giải ở dưới nhé !
Giải thích các bước giải:
Câu 4:
Cu không tác dụng với dd HCl ở nhiệt độ thường.
\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
a.\\
{n_{{H_2}}} = 0,25mol\\
\to {n_{Zn}} = {n_{{H_2}}} = 0,25mol\\
\to {m_{Zn}} = 16,25g\\
\to {m_{Cu}} = 13,75g
\end{array}\)
\(\begin{array}{l}
b.\\
\% {m_{Zn}} = \dfrac{{16,25}}{{30}} \times 100\% = 54,17\% \\
\% {m_{Cu}} = 100\% - 54,17\% = 45,83\%
\end{array}\)
\(\begin{array}{l}
c.\\
{n_{HCl}} = 2{n_{Zn}} = 0,5mol\\
\to {m_{HCl}} = 18,25g\\
\to C{\% _{HCl}} = \dfrac{{18,25}}{{200}} \times 100\% = 9,125\%
\end{array}\)
Câu 6:
\(\begin{array}{l}
Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}\\
{n_{Mn{O_2}}} = 0,9mol
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{HCl}} = 4{n_{Mn{O_2}}} = 3,6mol\\
\to {m_{HCl}} = 131,4g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{131,4}}{{20\% }} \times 100\% = 657g
\end{array}\)
\(\begin{array}{l}
b.\\
{n_{MnC{l_2}}} = {n_{Mn{O_2}}} = 0,9mol\\
\to {m_{MnC{l_2}}} = 113,4g\\
{m_{{\rm{dd}}}} = {m_{Mn{O_2}}} + {m_{{\rm{dd}}HCl}} - {m_{C{l_2}}} = 671,4g\\
\to C{\% _{MnC{l_2}}} = \dfrac{{113,4}}{{671,4}} \times 100\% = 16,89\%
\end{array}\)
\(\begin{array}{l}
c.\\
C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O\\
{n_{C{l_2}}} = {n_{Mn{O_2}}} = 0,9mol\\
\to {n_{NaOH}} = 2{n_{C{l_2}}} = 1,8mol\\
\to {n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,9mol\\
\to C{M_{NaOH}} = \dfrac{{1,8}}{{0,25}} = 7,2M\\
\to C{M_{NaCl}} = C{M_{NaClO}} = \dfrac{{0,9}}{{0,25}} = 3,6M
\end{array}\)
d,
\(\begin{array}{l}
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{FeC{l_3}}} = \dfrac{2}{3}{n_{C{l_2}}} = 0,6mol\\
\to {m_{FeC{l_3}}} = 97,5g\\
\to {m_{dd}} = 97,5 + 52,5 = 150g\\
\to C{\% _{FeC{l_3}}} = \dfrac{{97,5}}{{150}} \times 100\% = 65\%
\end{array}\)
